How did we get 807kJ from question a？how did we get 128kJ for b ？is there a step by step explanation？ 66. Using the bond energies in Table 7.3, determine the approximate enthalpy change for each ofthe following reactions:(a) Cl, (g) + 3F,(g)→ 2CIF, (g)(b) H,C=CH,(g) + H,(g) → H,CCH,(g)Solution(a) -807 kJ;EDDe-c + 4Dc-H + Du-H - Dc-c - 6Dc-H:= 611 + 4(415) + 436 -= -128 kJDH°ZDbonds formedbonds broken%3DC-C(b)345 - 6(415)| BondBond EnergyBondBond EnergyBondBond EnergyH-H436C-S260F-CI255H-C415C-CI330F-Br235H-N390C-Br275Si-Si230H-O464C-I240SI-P215H-F569N-N160Si-S225H-Si395N = N418Si-CI359H-P320N = N946Si-Br290H-S340N-O200Si-l215H-CI432N-F270P-P215H-Br370NAP210P-S230H-I295N-CI200P-CI330C-C345N-Br245P-Br270C = C611O-0140P-I215C = CO = 0S-S215837498C-N290O-F160S-CI250C = N615O-Si370S-Br215C = NO-P350CI-CI243891C-O350O-CI205CI-Br220C = 0741200CI-I210MacBook AirC = 01080F-F160Br-Br190C-F439F-Si540Br-l180C-Si360F-P489150C-P265F-S285

Please be noted that the value of enthalpy change for (b) is not - 128 kJ but it is found to be - 394 kJ.In the solution the C-C bond energy is written as 611 instead of 415 kJ. Bond energy or bond enthalpy is the energy which talks about the strength of the bond. This can be obtained from the bond dissociation enthalpies of all the bonds involved in the chemical reaction.66. Determine the bond enthalpy change for the following reactions: (a) Using the bond energies (bond enthalpies) values from Table 7.3, determine the enthalpy change: The heat of formation of ClF3 corresponds to the following reaction: Cl2(g) + 3 F2(g) → 2ClF3(g) In terms of bond enthalpies, this reaction corresponds to: 1) Breaking of one mole of Cl-Cl bonds and two moles of F-F bonds. 2) Formation of two ClF3 moles of Cl-F bonds. ∆H0 = ∑1∆Hbonds broken - ∑1∆Hbonds formed ∆H0 =(Cl-Cl)+3×(F-F) - 2×(3×(Cl-F)∆H0 =[243+(3×160)] kJ - [2×(3×255)] kJ∆H0 =723 kJ - 1530 kJ = - 807 kJ Hence, the enthalpy change for the above reaction is - 807 kJ. (b) Using the bond energies (bond enthalpies) values from Table 7.3, determine the enthalpy change: The heat of formation of ClF3 corresponds to the following reaction: CH2=CH2(g) + H2(g) → CH3-CH3(g) In terms of bond enthalpies, this reaction corresponds to: 1) Breaking of one C-C, 4 C-H and one H-H bonds. 2) Formation of one C-C and 6 C-H bonds. ∆H0 = ∑1∆Hbonds broken - ∑1∆Hbonds formed ∆H0 =[(C-C)+4×(C-H)+(H-H)] - [(C-C)+6(C-H)]∆H0 =[345+(4×415)+436] kJ - [345+(6×415)] kJ∆H0 =- 394 kJ Hence, the enthalpy change for the above reaction is - 394 kJ.